/*
 * @lc app=leetcode.cn id=117 lang=java
 *
 * [117] 填充每个节点的下一个右侧节点指针 II
 *
 * https://leetcode-cn.com/problems/populating-next-right-pointers-in-each-node-ii/description/
 *
 * algorithms
 * Medium (39.24%)
 * Likes:    73
 * Dislikes: 0
 * Total Accepted:    8K
 * Total Submissions: 19.8K
 * Testcase Example:  '{"$id":"1","left":{"$id":"2","left":{"$id":"3","left":null,"next":null,"right":null,"val":4},"next":null,"right":{"$id":"4","left":null,"next":null,"right":null,"val":5},"val":2},"next":null,"right":{"$id":"5","left":null,"next":null,"right":{"$id":"6","left":null,"next":null,"right":null,"val":7},"val":3},"val":1}'
 *
 * 给定一个二叉树
 * 
 * struct Node {
 * ⁠ int val;
 * ⁠ Node *left;
 * ⁠ Node *right;
 * ⁠ Node *next;
 * }
 * 
 * 填充它的每个 next 指针，让这个指针指向其下一个右侧节点。如果找不到下一个右侧节点，则将 next 指针设置为 NULL。
 * 
 * 初始状态下，所有 next 指针都被设置为 NULL。
 * 
 * 
 * 
 * 示例：
 * 
 * 
 * 
 * 
 * 输入：{"$id":"1","left":{"$id":"2","left":{"$id":"3","left":null,"next":null,"right":null,"val":4},"next":null,"right":{"$id":"4","left":null,"next":null,"right":null,"val":5},"val":2},"next":null,"right":{"$id":"5","left":null,"next":null,"right":{"$id":"6","left":null,"next":null,"right":null,"val":7},"val":3},"val":1}
 * 
 * 
 * 输出：{"$id":"1","left":{"$id":"2","left":{"$id":"3","left":null,"next":{"$id":"4","left":null,"next":{"$id":"5","left":null,"next":null,"right":null,"val":7},"right":null,"val":5},"right":null,"val":4},"next":{"$id":"6","left":null,"next":null,"right":{"$ref":"5"},"val":3},"right":{"$ref":"4"},"val":2},"next":null,"right":{"$ref":"6"},"val":1}
 * 
 * 解释：给定二叉树如图 A 所示，你的函数应该填充它的每个 next 指针，以指向其下一个右侧节点，如图 B 所示。
 * 
 * 
 * 
 * 提示：
 * 
 * 
 * 你只能使用常量级额外空间。
 * 使用递归解题也符合要求，本题中递归程序占用的栈空间不算做额外的空间复杂度。
 * 
 * 
 */
/*
// Definition for a Node.
class Node {
    public int val;
    public Node left;
    public Node right;
    public Node next;

    public Node() {}

    public Node(int _val,Node _left,Node _right,Node _next) {
        val = _val;
        left = _left;
        right = _right;
        next = _next;
    }
};
*/
class Solution {
    public Node connect(Node root) {
        if (root == null) {
            return null;
        }
        if (root.right != null) {
            connect(root.right, root.next);
        }
        if (root.left != null) {
            connect(root.left, root);
        }

        connect(root.right);
        connect(root.left);
        return root;
    }
    
    private void connect(Node start, Node nextPar) {
        while (nextPar != null) {
            if (nextPar.left != null && nextPar.left != start) {
                start.next = nextPar.left;
                break;
            }
            if (nextPar.right != null) {
                start.next = nextPar.right;
                break;
            }
            nextPar = nextPar.next;
        }
    }
}

